package com.dy.其他.有效的括号;

import java.util.HashMap;
import java.util.Stack;

/*
有效的括号
给定一个只包括 '('，')'，'{'，'}'，'['，']' 的字符串，判断字符串是否有效。

有效字符串需满足：

左括号必须用相同类型的右括号闭合。
左括号必须以正确的顺序闭合。
注意空字符串可被认为是有效字符串。

示例 1:

输入: "()"
输出: true
示例 2:

输入: "()[]{}"
输出: true
示例 3:

输入: "(]"
输出: false
示例 4:

输入: "([)]"
输出: false
示例 5:

输入: "{[]}"
输出: true
 */
public class Solution {
    public static boolean isValid(String s) {
        if(s.isEmpty()) return true;
        Stack<Character> stack = new Stack();
        for(Character c:s.toCharArray()){
            if(c=='('){
                stack.push(')');
            }
            else if(c=='{'){
                stack.push('}');

            }
            else if(c=='['){
                stack.push(']');
            }
            else if(stack.isEmpty()|| stack.pop()!=c){
                return false;
            }
        }
        return  stack.isEmpty();
    }

//用hashmap配对，每个字符判断是不是右半部分，是的话，栈pop，并比较左半部分，否则返回false，不是右半部分，入栈
    private HashMap<Character, Character> mappings;

    // Initialize hash map with mappings. This simply makes the code easier to read.
    public Solution() {
        this.mappings = new HashMap<Character, Character>();
        this.mappings.put(')', '(');
        this.mappings.put('}', '{');
        this.mappings.put(']', '[');
    }

    public boolean isValid2(String s) {

        // Initialize a stack to be used in the algorithm.
        Stack<Character> stack = new Stack<Character>();

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);

            // If the current character is a closing bracket.
            if (this.mappings.containsKey(c)) {

                // Get the top element of the stack. If the stack is empty, set a dummy value of '#'
                char topElement = stack.empty() ? '#' : stack.pop();

                // If the mapping for this bracket doesn't match the stack's top element, return false.
                if (topElement != this.mappings.get(c)) {
                    return false;
                }
            } else {
                // If it was an opening bracket, push to the stack.
                stack.push(c);
            }
        }

        // If the stack still contains elements, then it is an invalid expression.
        return stack.isEmpty();
    }
}
